Finding wire tensions in a symmetric hanging traffic light

Expert intro: Tension components in a symmetric hanging system are what make this traffic light stay at rest, and the 10° angle controls the force balance.

Detailed walkthrough

Identify the force balance

Because the traffic light is hanging motionless, the upward components of the two wire tensions must add up to the full weight of 2500 N. The setup is symmetric: both wires have the same length, and both make the same angle of 10° below the horizontal. That means the horizontal components cancel, while the vertical components support the load.

Each wire contributes the same vertical component, so we can write

$2T\sin(10^\circ)=2500$

if the angle is measured below the horizontal. This is the key relationship, because the vertical component of each tension is the part that resists gravity.

Solve for the tension in each wire

Rearrange the equation to isolate $T$:

$T=\frac{2500}{2\sin(10^\circ)}$

Using $\sin(10^\circ)\approx 0.17365$,

$T\approx \frac{2500}{2(0.17365)}\approx 7198\text{ N}$

So each wire has a tension of about

$\boxed{7.2\times 10^3\text{ N}}$

The large value is reasonable because a shallow angle means the wire must pull very hard to provide enough upward lift.

Why the geometry matters

The 10° angle is small, so most of the tension is horizontal rather than vertical. That is why the actual tension is much larger than 2500 N, even though the load itself is only 2500 N. When the wire is closer to horizontal, the sine value gets smaller and the tension must increase sharply.

For this reason, problems like this are solved by choosing the component that points upward and using equilibrium in the vertical direction. The horizontal forces do not disappear individually; they cancel because the arrangement is symmetric.[1][2]

Common checking step

A quick reasonableness check helps here: if the angle were 90° from the horizontal, each wire would be straight up and each tension would be only 1250 N. Since 10° is much flatter, the tension must be much larger than 1250 N. The calculated value near 7200 N fits that expectation.

💡 Pitfall guide

A frequent mistake in a 10° below-the-horizontal tension problem is using cosine instead of sine for the vertical component. The 10° angle is measured from the horizontal, so the vertical part of each tension is T sin(10°), not T cos(10°). Another trap is forgetting that there are two wires sharing the load, which leads students to set T sin(10°)=2500 and get an answer that is too large by a factor of 2. It also helps to keep the geometry consistent: the two horizontal components are equal and opposite, so they cancel, but they do not help support the weight. If a student draws the angle from the vertical instead of the horizontal, the entire component setup changes and the final number will be wrong even if the algebra is correct. A clean free-body diagram prevents most of these errors.

🔄 Real-world variant

If the same traffic light weighed 3000 N instead of 2500 N, with the wires still making 10° below the horizontal, the force balance would become 2T sin(10°) = 3000. That gives T = 3000 / [2 sin(10°)], so each wire would need a tension of about 8638 N. If the weight stayed 2500 N but the wires were raised to 15° below the horizontal, the equation would be 2T sin(15°) = 2500, which lowers the tension to about 4830 N. This shows how strongly the angle controls the wire tension: a larger angle gives a larger vertical component and therefore a smaller required tension. In a real engineering setting, that relationship is critical when choosing cable geometry for signs, lights, and suspension supports.

🔍 Related terms

force equilibrium in cables, tension component analysis, free-body diagram statics